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Call $f(n)=k$. I call $g(n)=\lceil\frac{2n+1}{3}\rceil$, and claim $f(n)=g(n)$ for all $n>5$. Also, $f(5)=3$. To prove this, I will need two lemmas: 1. $f(n+3)\ge f(n)+2$ 2. $f(n)\le g(n)$ Proof of lemma 1. Take any non-right angle ABC is an n-gon with f(n) right angles. There exists such a non-right angle since n>4. If we replace B with two points arbitrarily close to B on AB, BC respectively (call them D and G), and then draw a square DEFG away from the inside of the polygon, then by replacing ...ABC... by ...ADEFGC..., we have an (n+3)-gon with at least f(n)+2 right angles. Note that ...ADEFGC... is not self-intersecting in general since we can make square DEFG arbitrarily small. Hence $f(n+3)\ge f(n)+2$. Proof of lemma 2. An n-gon with f(n) right angles has sum of non-right angles $180(n-2)-90f(n)<360(n-f(n))$. Hence $f(n)<\frac{2n+4}{3}$, and $f(n)\le\lceil\frac{2n+1}{3}\rceil=g(n)$, as desired. According to lemma 2, $f(n)\le g(n)$ for $n=6$, $7$, $8$. That $f(n)=g(n)$ is actually achievable in these cases, and $f(5)=3$, can be easily checked. Now assume that $f(n)=g(n)$. Then lemma 2 says that $f(n+3)\le g(n+3)$. Also, lemma 1 says that $f(n+3)\ge f(n)+2=g(n)+2=g(n+3)$. Consequently, $f(n+3)=g(n+3)$, and we are done by induction on n.

i think the answer is [(2n+3)/3]

I shall show that when $n = 5$ we have $k = 3$ and when $n > 5$ we have $k = \left\lceil\frac{2n + 1}{3}\right\rceil$. Lemma 1: When $n = 5$ we have $k = 3$ Proof: First I shall construct an example where the number of interior right angles is $3$. Consider a square $ABCD$. Let $E$ be a point not in the interior of $ABCD$ such that $AED$ is an isosceles right triangle with hypotenuse $AD.$ Then the pentagon $ABCDE$ clearly has $3$ interior right angles as desired. Now, I shall show that there may not be $4$ interior right angles. Assume the contrary. Then since the sum of interior angles must be equal to $540$, we have that the non-right interior angle in the pentagon has angle measure of 180 degrees, contradiction. Similarly $5$ interior right angles are impossible so in this case $k = 3$ as desired. Lemma 2: For all integers $n \geq 5$, we have that $k \leq \left\lceil\frac{2n + 1}{3}\right\rceil$. Proof: Consider a $n$-gon with $k$ interior right angles. Then the sum of the non-right interior angles is $180(n - 2) - 90k$. But every such angle has angle measure less than 360 degrees so we have the inequality $180(n - 2) - 90k < 360(n - k)$ which yields the desired result. Lemma 3: When $n = 6$ we can construct a figure with $5$ interior right angles. Proof: Consider a square $ABCD$. Construct points $E, F, G, H$ all distinct from $A, B, C, D$ such that quadrilaterals $ABEF$ and $BCGH$ are squares. Then the hexagon $BEFDGH$ has $5$ interior right angles as desired. Lemma 4: When $n = 7$ we can construct a figure with $5$ interior right angles. Proof: Consider hexagon $BEFDGH$ from the $n = 6$ case. Let $M, N$ be the midpoints of segments $BE, BH$ respectively. Then heptagon $DGHNMEF$ has $5$ interior right angles as desired. Lemma 5: When $n = 8$ we can construct a figure with $6$ interior right angles. Proof: Consider hexagon $BEFDGH$ from the $n = 6$ case. Let $M$ be the midpoint of segment $EF$. Construct points $R, S$ not in the interior of the hexagon such that quadrilateral $FMRS$ is a square. Then octagon $DGHBEMRS$ has $6$ interior right angles as desired. Lemma 6: For any $n$-gon with $k$ interior right angles, we can construct an $(n + 3)$-gon with $k + 2$ interior right angles. Proof: Let the $n$-gon be denoted $A_{1}A_{2} \dots A_{n}$. Assume WLOG that $\angle{A_{1}A_{2}A_{3}} \ne 90$. Let $X$ be a point distinct from $A_2$ on segment $A_{1}A_{2}$ and let $Y$ be a point distinct from $A_2$ on segment $A_{3}A_{2}$. Construct points $V, W$ not in the interior of the $n$-gon such that quadrilateral $XYVW$ is a square. Then the $(n + 3)$-gon $A_{1}XWVYA_{3}A_{4} \dots A_n$ has $k + 2$ interior right angles as desired. By making the distances $A_{2}X$ and $A_{2}Y$ sufficiently small we can force the resulting $(n + 3)$-gon to not self-intersect and so we have the desired result. The combination of these lemmas clearly proves my initial claim.

$k_n = \left\lceil\frac{2n + 4}{3}\right\rceil -1$ is the answer. Using sum-of-angles formula, one gets this as upepr bound. I prove by induction that is also lower bound. For $5,6,7$ is easy. now I prove we can get $n+3$-gon with 2 more 90 angles than $n$-gon. For $n$-gon with $k_n$ 90 angles, there are two consecutive ones since $k_n > n/2$. If you change those two consecutive ones like in the figure, we are done! (Notice we can make the change very close to original figure, so that the polygon is still not self intersects!) (original polygon is right, result polygon is left)

I claim that $k_n = \left\lceil\frac{2n+1}{3}\right\rceil - 1$ UPPER BOUND: Consider a $n$-gin with $k$ right angles. Then the sum of the non-interior angles will be $180(n-2) - 90k$. Since every non-interior angle is less than 360, we have $180(n-2) - 90k < 360(n-k)$. Our desired conclusion follows from this. It is trivial to check our claim for $n \leq 8$. CONSTRUCTION: We will construct by induction. Suppose we have $k_n$ right angled $n$-gon. Let $A,B$ be points on the $n$-gon such that $\angle{A} = 90, \angle{B} < 90$. It is easy to see that such points exist. Now consider points $C,D$ such that $C$ is outside the $n$-gon and $D$ is on the $n$-gon such that their inclusion into the polygon will add $2$ more right angles. Obviously $2$ such points exist. Hence we are done by induction.

Let $f(n)$ be the maximum number of right angles for a polygon with $n$ vertices. We claim $$f(n) =\begin{cases} 3 &\text{ if } n=5 \\ 2m+1 & \text{ if } n=3m \\ 2m+1 &\text{ if } n=3m+1 \\ 2m+2 &\text{ if } n=3m+2.\end{cases}$$ Proof of Bound It is well-known that the sum of the internal angles in a polygon (be it non-convex) with $n$ vertices is $180(n-2)$. Every internal angle is less than $360^\circ$, so $$90f(n)+360(n-f(n)) > 180(n-2) \implies f(n) < \frac{2n+4}{3}.$$ Since $f(n)$ is an integer, this implies $f(n) \le \left\lfloor \frac{2n+3}{3} \right\rfloor$. Construction The idea is that by the above bound, we want the non-$90^\circ$ angles to be very close to $360^\circ$. We use different constructions based on $n\pmod 3$. Firstly, for $n=5$, it is easy to see that we actually cannot have $\lfloor \tfrac{2\cdot 5+3}{3} \rfloor = 4$ right angles, we can only have 3. For $n\ge 6$, the following constructions work (drawn in red): We start with a right angle at the wedge'', and draw two segments radially outward. Then mark points along the arc created, and erect squares on each of these segments. Connecting all the points gives the above shape for $n\equiv 0 \pmod3$. For $n\equiv 1 \pmod3$, we don't need any more right angles, so the filler segment is added. For $n\equiv 2 \pmod3$, we add two right angles in the wedge as shown (removing the central right angle), which adds a net of 1 more right angle, as needed.

Consider our exterior angles, which sum to $360^\circ$. Note that each right angle has an exterior angle of $90^\circ$, and each other angle has an exterior angle of less than $-180^\circ$ so an $n$-gon has $k$ right angles then $360-90k< -180(n-k)$ so $3k<2n+4$ meaning that $$k\le \left\lfloor \frac{2k+3}3 \right\rfloor$$ It is easy to check that for $n=5,6,7,8$ the corresponding $k_{\text{max}}$ are $3,5,5,6$. For higher $n$, we can do something funny like this to get $+3$ sides but $+2$ right angles.

Let $k(n)$ denote the maximum number of right angles in a polygon with $n$ vertices which is not self-intersecting. We claim that, $k(5)=3$ and $$k(n)=\lceil{\frac{2n+4}{3}}\rceil-1$$ for all $n\geq 6$. For $n=5$ it is clear that we cannot have $5$ right angles as the sum of internal angles in $540^\circ$. Further, we cannot have $4$ right angles as this implies that the other angle is $180^\circ$ which is a clear contradiction. It is clear that we can have 3 right angles - simply consider a rectangle with a slice cut off from one corner. Now, consider $n\geq 6$. We first deal with the bound. Say we have $k$ right angles in the polygon. Let $S$ be the sum of internal angles of this $n$ vertex polygon. We count $S$ in two ways. $$180(n-2)=S < 90k + 360(n-k) \implies k < \frac{2n+4}{3}$$ From this, it follows that, $$k \leq \lceil{\frac{2n+4}{3}\rceil}-1$$ as desired. Now, we deal with the difficult part of this problem - the construction. When $n>5$ and $n\equiv 0 \pmod{3}$ ($n=3a+6$ for $a\geq 0$), we simply consider the following construction. We construct a petal - which is a L shaped structure, with 3 edges and two right angles between them. Further, we draw $n$ of these, and connect the top most edge of the first petal and the bottom edge of the last petal to form a rectangular structure. This requires $3a+6$ edges and forms $2a+5$ right angles. Thus, for all $n>5$ such that $n\equiv 0 \pmod{3}$ we can achieve the desired bound with this construction. For $n\equiv 1 \pmod{3}$ we simply break the edge from the end of the last petal into two edges which have no right angles between them. This achieves the desired bound. When $n\equiv 2 \pmod{3}$ we draw an additional right angle before connecting to the bottom petal adding one more right angle and one more side. Thus, this bound is achievable in all cases and we are done. This proof is very hard to explain without plenty of diagrams - I add the cases $n=15,16,17$ to make this clear.

i think it's nontrivial that the sum of internal angles is $180(n-2)$ for a general $n$-gon, i gave the problem to orz jordan lefkowitz and he hasn't solved it in 5 seconds so that proves my point (i could be wrong tho) I claim the sum of internal angles of the polygon is $180(n-2)$. It suffices to show that some diagonal of the polygon splits in induct downward to the trivial $3$-gon. First I prove there is a convex angle $\angle BAC$. Take a point $A$ on the convex hull; the two sides $AB$ and $AC$ from point $A$ form a convex angle; it suffices to show this angle is on the interior. If not, then the reflex angle $\angle BAC$ is on the interior; however some point $P$ outside the convex hull with $AP=\epsilon$ which is inside the reflex angle must exist. Thus $P$ is on the interior of the polygon but on the exterior of the convex hull, a contradiction. Now I focus on this angle $\angle BAC$. If there does not exist a point inside $\triangle ABC$ then $BC$ is the desired diagonal. Otherwise choose the point $P$ inside the triangle with the farthest distance from $BC$. Evidently no side of the polygon can intersect $AP$, hence $AP$ divides the polygon. Remark: There's probably some sketchy things regarding three vertices of the polygon being collinear, but if $0^{\circ}$ angles are allowed it is okay. (Hi continuity) Now we have proven the claim. Now notice $$\sum(180^{\circ}-\alpha)=360^{\circ}$$ where we sum over all angles $\alpha<360^{\circ}$. Hence if there are $k$ right angles, we have $$360=90k+\sum(180-\alpha)>90k-180(n-k)\implies 4>k-2(n-k)=3k-2n\implies k<\frac{2n+4}{3}.$$ Now it suffices to find constructions for: $(n,k)=(3a,2a+1)$. $(n,k)=(3a+1,2a+1)$. $(n,k)=(3a+2,2a+2)$. The answer for $n=5$ is actually $k=3$; we cannot have $k=4$ as the non-right-angle is $180^{\circ}$. The construction: a square with a corner removed. Hence we assume $a\ge 2$. Here is an image of the construction: